Let $\omega=xdx\wedge dz-zdy\wedge dz$ and $X=f\frac{\partial}{\partial x}+g\frac{\partial}{\partial y}$, for some $f,g\in C^\infty(\mathbb{R}^3).$
I want to compute $\omega(X,\frac{\partial}{\partial z})$.
I have the solutions available and still can't understand the formulas used, since the only thing I found was $\omega(gX)=g\omega(X)$, for $\omega$ a form of degree 1. (which probably is not useful here)
The first step I have is $\omega(X,\frac{\partial}{\partial z})=f\omega(\frac{\partial}{\partial x},\frac{\partial}{\partial z})+g\omega(\frac{\partial}{\partial y},\frac{\partial}{\partial z})$.
Then, the second and last step I have is $f\omega(\frac{\partial}{\partial x},\frac{\partial}{\partial z})+g\omega(\frac{\partial}{\partial y},\frac{\partial}{\partial z})=fxdx\wedge dz(\frac{\partial}{\partial x},\frac{\partial}{\partial z})-fzdy\wedge dz(\frac{\partial}{\partial x},\frac{\partial}{\partial z})+gxdx\wedge dz(\frac{\partial}{\partial y},\frac{\partial}{\partial z})-gzdy\wedge dz(\frac{\partial}{\partial y},\frac{\partial}{\partial z})$
Which is then equal to $fx-gz$.
This part of the syllabus has some very complicated notation for me and I am struggling to understand what it is I really am doing, so I ask you to kindly be soft with your explanation.
PS: Am I computing $\omega$ $\mathbf{of}$ $(X,\frac{\partial}{\partial z})$? Is it $\omega$ $\mathbf{times}$ $(X,\frac{\partial}{\partial z})$?